Wooden Sticks（区间覆盖）

Wooden Sticks
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Description
There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:

(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l’ and weight w’ if l<=l’ and w<=w’. Otherwise, it will need 1 minute for setup.

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, …, ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output
The output should contain the minimum setup time in minutes, one per line.

Sample Input
3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1

Sample Output
2
1
3

题意：首行给出T，代表T组数据。每组数据第一行给出N，接着一行有N组区间坐标，共N*2个数据，如果后一个区间的左右坐标均大于等于前一个区间的左右坐标，那么就不计算时间花费，否则就要花费一分钟，输出处理完所有区间需要多少分钟

思路：首先按照左坐标升序排列，如果左坐标相等就按照右坐标升序排列。处理过的区间要进行标记。

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<queue>
using namespace std;
struct node
{
    int l;
    int w;
    int vis;//初始化为0，用过标记为1
} num[5005];
bool cmp(node x,node y)
{
    if(x.l==y.l)
        return x.w<y.w;
    return x.l<y.l;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int N;
        scanf("%d",&N);
        for(int i=0; i<N; i++)
        {
            scanf("%d%d",&num[i].l,&num[i].w);
            num[i].vis=0;
        }
        sort(num,num+N,cmp);
        int sum=0;//标记的数量
        int result=0;//需要的时间
        int left,right;//记录左右边界
        while(sum!=N)//全标记以前一直循环
        {
            for(int i=0; i<N; i++)
            {
                if(num[i].vis==0)
                {
                    left=num[i].l;
                    right=num[i].w;
                    sum++;
                    result++;
                    num[i].vis=1;
                    break;
                }
            }
            for(int i=0; i<N; i++)
            {
                if(num[i].vis==0&&num[i].l>=left&&num[i].w>=right)
                {
                    sum++;
                    left=num[i].l;
                    right=num[i].w;
                    num[i].vis=1;
                }
            }
        }
        printf("%d\n",result);
    }
    return 0;
}